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Topic: digital root (Read 1589 times) |
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Christine
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digital root
« on: Jan 8th, 2013, 12:34pm » |
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Why is the digital root of the product of twin primes, other than (3,5), 8?
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towr
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Re: digital root
« Reply #1 on: Jan 8th, 2013, 10:52pm » |
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Because the digital root of n is 1+ (n-1) mod 9
And twin primes greater than (3,5) are of the form (6n-1,6n+1). So their product is 36n2 - 1 = -1 (mod 9).
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Christine
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Re: digital root
« Reply #2 on: Jan 9th, 2013, 10:47am » |
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on Jan 8th, 2013, 10:52pm, towr wrote:Because the digital root of n is 1+ (n-1) mod 9
And twin primes greater than (3,5) are of the form (6n-1,6n+1). So their product is 36n2 - 1 = -1 (mod 9). |
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I see it now,
(6n - 1)(6n + 1) = 36*n^2 - 1
because we know that the digital root of a square is 1, 4, 7, or 9
so 36*n^2 mod 9 - 1 = -1
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towr
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Re: digital root
« Reply #3 on: Jan 9th, 2013, 11:59am » |
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I suppose that's another way, if that's a well-known property of digital roots. It's implied by the first line of my post, but we can also apply that directly to drop the first term (since 36 is a multiple of 9).
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tsitut
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Re: digital root
« Reply #4 on: Feb 20th, 2013, 12:13pm » |
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hm right XD I feel sh*tty I couldn't find the right answer without looking :/
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